Question

What is the exact value of cos((19π)/(12))

Answer 1

0.2588190451 ;)

Answer 2

thats wrong..

Answer 3

Try this:\[\cos \frac{ 19 \pi }{ 12 }=\cos \left( \frac{ 16\pi }{ 12 }+\frac{ 3\pi }{ 12 } \right)\]
Now simplify the fractions. You wil get numbers we all know and love ;)
Then use the formula for cos(a+b).

Answer 4

nope that aint wrong ;)

Answer 5

@itsmylife: it's wrong because you left out infinitely many decimals...
It is just a (good) approximation.

Answer 6

\[\frac{ \sqrt{6}+\sqrt{2} }{ 4 }\]

Answer 7

is that correct?

Answer 8

I got \[\frac{ \sqrt{6}-\sqrt{2} }{ 4 }\]

Answer 9

well she asked me exact value i used calculator and gave her what she wanted ;)

Answer 10

can you explain ?
& @itsmylife but thats approximent. exact value is usually always fractions.

Answer 11

alright m gonna give ya fraction ;)

Answer 12

647/2500 ;) this is way too exact

Answer 13

cos(a+b)=cos(a)cos(b)-sin(a)sin(b)

Here we have \[a=\frac{ 16\pi }{ 12 }=\frac{ 4 }{ 3 }\pi\]and \[b=\frac{ 3\pi }{ 12 }=\frac{ 1 }{ 4 }\pi\]
Substitute these values:\[\cos \frac{ 4 }{ 3 }\pi \cos \frac{ 1 }{ 4 }\pi-\sin \frac{ 4 }{ 3 }\pi \sin \frac{ 1 }{ 4 }\pi=\]\[-\frac{ 1 }{ 2 }\cdot \frac{ 1 }{ 2 }\sqrt{2}--\frac{ 1 }{ 2 }\sqrt{3}\cdot \frac{ 1 }{ 2 }\sqrt{2}=\frac{ 1 }{ 4 }\sqrt{6}-\frac{ 1 }{ 4 }\sqrt{2}=\frac{ \sqrt{6}-\sqrt{2} }{ 4 }\]

Answer 14

@Lauren01: seems like you missed a little "-"sign somewhere...